Generator Efficiency

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Here several useful generator equations are derived.

Electrical Efficiency

Often, because maximum power transfer occurrences when impedance is match, there are designs where the generator load equals the internal source resistance. Sure maximum possible power is transferred, but 1/2 of the mechanical input becomes heat inside the generator. Instead, oversize the generator, and use a fraction of the possible power to keep efficiency.

Electrical efficiency as a fraction input mechanical power that is available as electricity a the output terminals. Of course the voltage of the available power might not be optimal - but thats what switching power supplies are for.

Rl = load resistance
Rm = motor series (internal) resistance

E = Pout / (Pout + Ploss)   (efficiency)

Pout = I^2 * Rl and  Ploss = I^2 * Rm thus

E = Rl / (Rl+Rm)

let Rl = k * Rm then  (define the load resistance as a fraction of the motor series resistance)

E = k/(1+k)

Thus to keep the efficiency high, then keep the load resistance
higher than the motor source resistance, or:

Rl >= Rm * E/(1-E)

some results:
k = 4 -> 80% 
k = 9 -> 90%

Power Efficiency

For a load that is a multiple of the generator resistance, a fraction of the maximum possible generator output power is obtained. If the load equals the internal generator resistance, then the maximum output power (Pmax) is obtained (but 1/2 of the input power becomes heat.) Here we can determine the electrical efficiency that results by over sizing the generation

Lets define the power efficiency be:

Pe = Pout / Pmax  (meaning take Pout from a generate that can supply Pmax)

some math...

Pe = 4k/(1+k)^2

some more math:
let b = 4/Pe - 2;
then 
k = 0.5 * (-b + sqrt(b^2 - 4))

64% -> k = 4   (80% electrical efficiency)
36% -> k = 9   (90% electrical efficiency)

Thus using a generator at 36% of its rated loads results 
in 90% of the mechanical energy being available as electrical energy.


Maximum Power

Its also possible to size a generator based on two fundamental properties of the generator. Here we derive the maximum generator power as a function of rotational velocity omega.

For a generator / motor:

omega = Ks * Vemf    (Vemf is back emf, and Ks is the speed constant)
or Vemf = omega/Ks

Vm = Vemf - I * Rm  (Vm is voltage at terminals)

I = Vm / Rl          
let Rl = Rm for max output power

then Vm = Vemf - (Vm/Rm)*Rm
or
Vm = omega/Ks - Vm

so at max output power
Vm = omega/2Ks

output power is V^2/R so

Pmax = Vm^2 / Rm 
or
Pmax = omega^2 / 4 Rm Ks^2

lets define the power constant as 
 
Pk = 1/4RmKs^2   

so Pmax = Pk * omega^2
For example:
A generator might produce 70 volts at 2500 rpm.  The internal resistance is 7.7 ohms.

ks = 2500/70 = 35.7 (rpm per volt) 

Pk  = 25e-6   units are v^2/(ohm rpm^2)
at 1000 rpm maximum power is 25.5 watts
at 2000 rpm 101.8 watts.


A 2nd example is this 490g outrunner motor:

Rs = 0.03, ks = 295 rpm per volt
Pk = 1 / 4*0.03*295^2 = 96e-6 
at 1000 rpm, max power is 95 watts

3rd example : Himax HC6332-230 Outrunner:

Rs = 0.043, ks = 230 rpm per volt
Pk = 109e-6
at 1000 rpm, max power is 109 watts

Powermax EC30-48V :

Rs = .386, ks = 348 
Pk = 5.3e-6
at its peak speed of 25,000 rpm - this little motor could produce 3300 watts 
(and overheat in a few millisconds!)  thus this is a great high speed motor 
to use at fraction of its possible power.